Lời giải:
Ta thấy:
\(a^2+b^2+c^2-(ab+bc+ac)=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\geq 0, \forall a,b,c\in\mathbb{R}\)
\(\Rightarrow a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow 2(a^2+b^2+c^2)\geq 2(ab+bc+ac)\)
\(\Rightarrow 3(a^2+b^2+c^2)\geq (a+b+c)^2\)
\(\Rightarrow a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}=\frac{1}{3}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Có: \(a^2+b^2+c^2\)\(\ge\)\(\frac{1}{3}\)
⇒\(3a^2+3b^2+3c^2\ge1\)
⇒\(3a^2+3b^2+3c^2\ge\left(a^2+b^2+c^2\right)^2\)(vì a+b+c=1)
\(3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2ac+2bc\)
⇒\(2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)
\(2a^2-2bc+2b^2-2bc+2c^2-2ac\ge0\)
\(\left(a^2-2ab+b^2\right)++\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\)
⇒\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
vì \(\left(a-b\right)^2\ge0\) với \(\forall\) a,b
\(\left(b-c\right)^2\ge0\) với \(\forall\)b,c
\(\left(c-a\right)^2\ge0\)với \(\forall\)c,a
⇒\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
luôn đúng
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